What is the integral of sin^{-5}(y)cos^7(y) ?

The integral of sin^{-5}(y)cos^7(y) is -1/4 csc^4(y)+3/2 csc^2(y)+3ln|sin(y)|-1/2 sin^2(y)+C

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integral of sin^{-5}(y)cos^7(y)

\int sin^{- 5} (y) cos^{7} (y) d y

- \frac{1}{4} csc^{4} (y) + \frac{3}{2} csc^{2} (y) + 3 ln ∣ sin (y) ∣ - \frac{1}{2} sin^{2} (y) + C

Solution steps

\int sin^{- 5} (y) cos^{7} (y) d y

Rewrite using trig identities

= \int (1 - sin^{2} (y))^{3} cos (y) sin^{- 5} (y) d y

Apply u-substitution

: \int \frac{( 1 - u ^{2} ) ^{3}}{u ^{5}} d u

= \int \frac{( 1 - u ^{2} ) ^{3}}{u ^{5}} d u

Expand

\frac{( 1 - u ^{2} ) ^{3}}{u ^{5}} : \frac{1}{u ^{5}} - \frac{3}{u ^{3}} + \frac{3}{u} - u

= \int \frac{1}{u ^{5}} - \frac{3}{u ^{3}} + \frac{3}{u} - u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{5}} d u - \int \frac{3}{u ^{3}} d u + \int \frac{3}{u} d u - \int u d u

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

\int \frac{3}{u ^{3}} d u = - \frac{3}{2 u ^{2}}

\int \frac{3}{u} d u = 3 ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= - \frac{1}{4 u ^{4}} - (- \frac{3}{2 u ^{2}}) + 3 ln ∣ u ∣ - \frac{u ^{2}}{2}

Substitute back

u = sin (y)

= - \frac{1}{4 sin ^{4} ( y )} - (- \frac{3}{2 sin ^{2} ( y )}) + 3 ln ∣ sin (y) ∣ - \frac{sin ^{2} ( y )}{2}

Simplify

- \frac{1}{4 sin ^{4} ( y )} - (- \frac{3}{2 sin ^{2} ( y )}) + 3 ln ∣ sin (y) ∣ - \frac{sin ^{2} ( y )}{2} : - \frac{1}{4} csc^{4} (y) + \frac{3}{2} csc^{2} (y) + 3 ln ∣ sin (y) ∣ - \frac{1}{2} sin^{2} (y)

= - \frac{1}{4} csc^{4} (y) + \frac{3}{2} csc^{2} (y) + 3 ln ∣ sin (y) ∣ - \frac{1}{2} sin^{2} (y)

Add a constant to the solution

= - \frac{1}{4} csc^{4} (y) + \frac{3}{2} csc^{2} (y) + 3 ln ∣ sin (y) ∣ - \frac{1}{2} sin^{2} (y) + C

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What is the integral of sin^{-5}(y)cos^7(y) ?
The integral of sin^{-5}(y)cos^7(y) is -1/4 csc^4(y)+3/2 csc^2(y)+3ln|sin(y)|-1/2 sin^2(y)+C